# extrema on a closed curve in $\mathbb R^3$

Let $S$ be the surface in $\mathbb R^3$ given by the equation $z=\frac12 x^2+\frac12 y^2$ and let $R\subset R^3$ be the surface given by $y^2+z^2=3$ ($R$ is the boundary of the infinite cylinder around the $x$-axis). Let $C$ be the closed curve defined as the intersection between $S$ and $R$. I want to find the maximum and minimum of $f(x,y,z)=x^2+y^2+(z-1)^2$ on $C$.

How can I do this? I've tried using the Lagrange multipliers, but it is quite a lot of work.

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Can you use the fact that, on the surface $S$, $2z = x^2+y^2$? Then, on the surface, $f(x,y,z) = z+(z-1)^2$, which you can easily maximize wrt $z$. You still need a way to use the second constraint, but I'm not quite sure what the second constraint is. What do you mean by "the surface given by $y^2+z^2$"? –  James Fennell Jul 28 '12 at 10:25
I made a typo, thank you for pointing it out –  kevin Jul 28 '12 at 10:27
If you succeed in finding a parametrization $c(t)$ of the intersection curve, then $f\circ c$ is a function $\mathbb{R}\rightarrow \mathbb{R}$. –  user20266 Jul 28 '12 at 11:11

James' observation simplifies the problem considerably: In all points of $S$, whence of $C$, we have $$f(x,y,z)=z^2-z+1=\Bigl(z-{1\over2}\Bigr)^2+{3\over4}\ .$$ Therefore we only have to control the value of $z$ along $C$.

The curve $C$ is a sling which encircles the $z$-axis in the $z>0$ part of ${\mathbb R}^3$. As $z=\sqrt{3-y^2}$ the value $z$ is maximal on $C$ when $y=0$, and we have $z_{\rm max}=\sqrt{3}$. Similarly, the minimal value of $z$ will be assumed in the plane $x=0$. Here we have $2z=y^2=3-z^2$; so $z_{\min}=1>{1\over2}$.

It follows that $f\restriction C$ is maximal in the two points $(\pm x_0 ,0,\sqrt{3})$, where $x_0$ is determined by the equation $2z=x^2$, and $f\restriction C$ is minimal in the two points $(0,\pm\sqrt{2},1)$.

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A smarter approach is to use the constraints to simplify the target function. From the second you'll get $f=x^2-2z+\mathrm{const}$, and then from the first $f=-y^2+\mathrm{const2}$. This clears up things a bit, especially with the form of the second constraint.